L35 : Numerical lens - Ray Optics, Physics, Class 12 Video Lecture

hello friends this video on ray optics
part 35 is brought to you by exam feel
calm no hope here from exam please make
sure that you have washed all the videos
to the pad 34 before going ahead with
part 35 so now that we have discussed
enough about lens it is we are good to
go with some problems on refraction by
lens so let us look at problem number 11
the problem says an object of size 3
centimeter is placed 14 centimeter in
front of a concave lens of focal length
21 centimeters describe the image
produced by the lens what happens if the
object is moved further away from the
lens so let us draw a concave lens which
would look somewhat like this we have an
object whose size is given as 3
centimeters so h/o has 3 centimeters it
is placed 14 centimeters in front of a
concave lens that means the object
distance is 14 centimeters since the
object lies somewhere here so this U is
negative focal length of the concave
lens is 21 centimeters for a concave
lens the focal length is also negative
right so we have to calculate or we have
to describe the image produced by the
lens now applying the thin lens formula
now why are we applying the thin lens
formula here because the problem here
doesn't talk about the radius of
curvature of both those temple surfaces
since it is not talking about the radius
of curvature that means it is assuming
that the lens is thin enough to ignore
the different values of the radii of
curvature so we will apply the thin lens
formula here which is 1 by V minus 1 by
U is equal to 1 by F now with the help
of this formula we can say 1 by V minus
1 by minus 14 is equal to 1 by minus 21
so solving this we get V is equal to
minus 42 by 5 which is equal to minus
eight point four centimeters so this
would be the location of the image this
is where the image will be formed but we
need to describe the nature of the image
as well for which we need to calculate
the magnification magnification is
nothing but height of the image by
height of the object and it is also
equal to the image distance by object
distance so we can write height of image
by height of object as 3 centimeters is
equal to V is minus eight point four
divided by U is minus 14 so from this we
get height of the image is equal to
eight point four divided by 14 into
three which comes out to be one point
eight centimeters so this shows that the
image is erect so since the image is
erect that means what do we get what
other information do we get the image is
erect so the image is also virtual and
what about the size of the image object
was three centimeters and the images 1
point 8 centimeters so the image is also
diminished and the image is located
eight point four centimeters from the
optical centre of the lens what happens
if the object is moved away is moved
further away from the lens now if the
object is moved further away the value
of U keeps increasing
now as you keeps increasing what happens
V keeps decreasing and as a result the
image approaches towards the optical
centre this is what we have described in
case of mirror also right as the object
keeps moving away from the lens the
image keeps moving towards the lens
right that is how the variation takes
place okay so I hope it is clear to you
now how do we approach problems related
to lens okay so let us go ahead to the
next problem which says double convex
lenses
to be manufactured from a glass of
refractive index 1.55 with both faces of
the same radius of curvature what is the
radius of curvature required if the
focal length is to be 20 centimeters it
says that you have got two spherical
surfaces that you have to make a double
convex lens with that it is given that
the radii of curvature of both the lens
are same that means R 1 and R 2 both are
equal to R and the lens which you will
form the focal length of that lens
should be 20 centimeters so you have to
calculate the radius of curvature
required for that and it is also given
that the refractive index of the
material that is glass is 1.5 v so in
this case we will apply the lens maker
formula why is this formula known as
lens maker formula because this formula
actually tells us how is the lens made
so it calls it so it is known as lens
maker formula because it tells us it
talks about the radii of curvature of
the surfaces which makes the lens that
is why it is called lens maker formula
so here we will apply lens make a
formula which says 1 by F is equal to MU
minus 1 into 1 by r1 minus 1 by r2 right
so using this 1 by 20 is equal to 1
point 5 5 minus 1 into 1 by ad minus
what is R 2 and this is R 1 is positive
and R 2 is going to be negative so this
will be 1 by ad plus 1 by r because if R
1 is R then R 2 will remind the Sun
because it is going to be a on mix lens
right therefore we say 1 by 20 is equal
to 0.55 into 2 by r or r is equal to 0.5
I 5 into 2 into 20 which comes out to be
22 centimeters so this would be the
radius of curvature required to make
this double convex lenss thank you
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